Apparent Power, KVA?

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Re: Apparent Power, KVA?

Ok, ok, maybe fiction was a little too metaphysical. . .
the " exact " equivalent circuit comprises 6 or more circuit elements. . .
Bold is my emphasis - Al
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Gary,

The value of the mathematical model is undeniably useful for teasing the various electromagnetic and physical phenomena apart for the purpose of conceptualization. . .but to look at the transformer eq. ckt. and think that it provides an arguement that apparent power or the current is all that is necessary to quantify a transformer is over simplification. The single impedance of a real transformer is deliciously complex.

There is no "exact" equivalent circuit.

Maybe some nice approximations, that include a post doctoral in math. . .
 
Re: Apparent Power, KVA?

Al, Please explain what you mean by the "single impedance" of a xfrmr.

I didn't name the "exact" equivalent. Some of the old heads did decades ago. To make the approximate circuit you move the two shunt elements up front and then combine the primary and reflected secondary impedances. Easier math.

Of course there is more to xfrmr ratings than the KVA rating. My original question was "Why do we include voltage in the KVA rating since voltage is essentially constant in power applications?" Someone said voltage has nothing to do with it, but I argued that core loss is a strong function of primary voltage, and this discussion ensued. Very interesting though.

Certainly, since the magnetics are so nonlinear, this small signal model will not be perfect. But, if the parameters are optimized for expected load and temperature, it should provide a reasonable prediction of phase shifts, losses, currents, voltage drops, etc.

Memory fades, but I think we had to characterize xfmrs with this model in the AC Machines lab back in the dark ages.

I am sure that there are non-linear models available now which are highly accurate over a wide range of conditions.
 
Re: Apparent Power, KVA?

Gary, I was in the middle of the calculator revolution. Three techs and I built the breadboard for the hp-35 in 1971--all 100 pounds of it. We did find some bugs which were corrected before committing to silicon.

Some time later, one of the slide rule companies visited, pulled out an hp-35, and said, "We want one of those!" Unfortunately, we had our hands full and could not help them.

Nowadays, you can buy scientific calculators in the grocery store!
 
Re: Apparent Power, KVA?

The discussion only about a transformer is a red herring. The transformer is a foil to get at the real discussion which is: "why KVA?" as asked at the outset.

Substitute a motor instead, . . .motor - transformer. . .either one has leads for the energy to be supplied through. Record a real time data track for current & voltage and the impedance can be represented. This is the "one impedance."
 
Re: Apparent Power, KVA?

Posted by Rattus
Certainly, since the magnetics are so nonlinear, this small signal model will not be perfect. But, if the parameters are optimized for expected load and temperature, it should provide a reasonable prediction of phase shifts, losses, currents, voltage drops, etc.
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. . .I think we had to characterize xfmrs with this model in the AC Machines lab back in the dark ages.
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I'm sure you did. The phase shift is the important thing to remember. In a real world motor or a transformer, the Apparant Power will always be less than the Total Power, the KVA.
 
Re: Apparent Power, KVA?

Okay Al, but I have a problem with characterizing a xfrmr or motor as a single impedance which changes with loading in either case. I would think you would want to know the losses, phase shifts, etc., to fully characterize the device.

Should be no great problem to determine this impedance with instruments, but what do you do with it other than use it as part of your load.

Then again, one can predict this single impedance with the xfrmr model and compare it to your measured values. It is quite easy with computer simulations. Don't even have to have a slide rule or understand trig!
 
Re: Apparent Power, KVA?

Posted by Rattus I have a problem with characterizing a xfrmr or motor as a single impedance which changes with loading in either case. I would think you would want to know the losses, phase shifts, etc., to fully characterize the device.
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Well, remember, that one impedance has both magnitude and direction. . .it is a vector quantity, highly nonlinear, as you say, but at any instant in time, a vector of given magnitude and direction.

When a sinusoidal voltage (represented as a vector) is applied to it, it will result in a phase shifted vector current. All the information is there. So, what's the problem. . .what's missing?

Edit for spelling.

[ December 31, 2004, 07:48 AM: Message edited by: al hildenbrand ]
 
Re: Apparent Power, KVA?

Al,

What is wrong is that your single impedance only begins to characterize the xfrmr. It is only useful for a fixed secondary load, and it tells you only the magnitude and phase of the primary current.

And a single impedance will not allow you to calculate regulation, and it will not separate the load power from the losses.

Now if you want to ignore the primary inductance and core loss elements, then you have resistance and inductance in series with the reflected load which is an approximation to the approximate circuit, and even that is useful.
 
Re: Apparent Power, KVA?

Posted by Rattus
And a single impedance will not allow you to calculate regulation, and it will not separate the load power from the losses.
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On the contrary, the actual observed impedance (single impedance) which is nonlinear, and even discontinuous (depending upon the physical construction of the apparatus), which is both real and reactive, i.e., complex, establishes the record that the results shown by models must fit.
Posted by Rattus
What is wrong is that your single impedance only begins to characterize the xfrmr.
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I'm hanging on to the single impedance, not because it characterizes , but, because it IS the apparatus.

From the observed data gathered at the apparatus input terminals, using energy supply of various steady or changing voltage and current and source impedance, the models are constructed that provide the insight that you are talking about. . .the models that apply under narrower conditions, such as, but not limited to, an energy source that is the "Infinite Bus" for purposes of calculation, no load, full load, etc.
 
Re: Apparent Power, KVA?

Al,

Please give me an example. I am not following you with this single impedance thing.

Explain to me how you can calculate regulation, efficiency, and phase shift with a single impedance.

In my mind, all impedances are linear. In this case we are applying a linear model to a non-linear device which is as we agree an approximation, but then much of what we do in engineering is an approximation. We just need to know the degree of error involved.
 
Re: Apparent Power, KVA?

Rattus posted:
In my mind, all impedances are linear.
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To go back to Charlie B.'s opening sentence:
It?s more complicated than that.
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The red herring is in the room when you ask:
calculate regulation, efficiency, and phase shift
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as we need to look at two very different outputs. . .the secondary of the xfmr or the shaft of the motor. While interesting and useful, that is off the question.

The answer to your question (Why KVA?) concerns what the Source "sees." It sees the input terminals of the apparatus.
 
Re: Apparent Power, KVA?

Al , Rattus

Personally, I think you can create an exact transformer model taking into account each major imperfection in turn. Since we are NOT dealing with an Ideal transformer then there will be no perfect model. That said, (the more the model is simplified the less accurate the results, and errors can be multiplied.).. transformer models are often more complex than needed in order to get good results. One of the chief complaints I have heard is that the excitation branch has a very small current compared to the load current just adds another node to the circuit being analyzed making the circuit more complex than necessary.
In fact, it is so small that under normal circumstances it causes a completely negligible voltage drop in Rp and Xp.
-Eddy current losses are resistive heating losses in the core and are proportional to the square of the voltage;
- Hysteresis losses are associated with the rearrangement of magnetic domains in the core during each half cycle. They are a complex, non-linear function of the voltage applied to the transformer.
- Leakage Fluxes for Lp and Ls escape the core and pass through only one winding of the transformer, these escaped fluxes create "Self-inductance" in the primary and secondary and their effects MUST be accounted for.
The idea of lumping everything into one impedance sounds too good to be true and after working with these things for quite awhile now I can not see how it is possible to equate all the parameters of a transformer to just one impedance and have any idea of what is really going on inside.


Gary
 
Re: Apparent Power, KVA?

Posted by Gary:
The idea of lumping everything into one impedance sounds too good to be true
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"Too good to be true". . . .no, too messy to be manipulated, it seems to me.

What do you see when you are away from the paper / monitor and are looking at an apparatus? A physical apparatus? I suspect that the various aspects of the various equivalent circuits for this apparatus inform your perspective.

It sounds like the physical apparatus is the red herring for you, no?
Edit - typo

[ December 31, 2004, 03:22 PM: Message edited by: al hildenbrand ]
 
Re: Apparent Power, KVA?

Al
I seem to have struck a stridant chord here and was not my intent. Your reply was that the equivalent circuit was 'Fiction' and not a close approximation, but on the contrary it is a very close approxmation and a good one.
From what you have written about the 'single impedance' to me means you must have a load connected to the transformer before you can determine what the impedance is.
By a mathematical model it gives one the opportunity to see what the impedance is before the load is applied and how it will affect the transformer and how to best match the transformer to the job and at various loads and as rattus pointed out transformation regulation comes from this too.
Messy ? I guess it depends, but a good approximation circuit of a transformer model can be reduced to basically a 'T' network and worked from there,.. so as far as what I perceive from the model either looking at the transformer, 'The Apparatus', or the model circuit is the same, the end result is hopefully the same, a safe efficient operation, isn't this what we are all about.

Gary
 
Re: Apparent Power, KVA?

Gary,
g1matrix posted December 31, 2004 06:58 PM
Your reply was that the equivalent circuit was 'Fiction' and not a close approximation
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al hildenbrand posted December 28, 2004 06:27 AM
You did not hear from my words that the equivalent circuit is "useless", rather that it is a fiction, something that is made up that models the behaviour of the real thing within acceptable error ranges.
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Say what?
 
Re: Apparent Power, KVA?

OK Gents, here it is:

Xformer.gif


In this circuit, Rp and Xp, are the primary resistance and primary leakage reactance which are both undesirable.

Lp is the primary inductance or magnetizing inductance. Rc is an equivalent resistance which accounts for the core loss.

Rs', Xs', and Zload' are the reflected secondary impedances. They have been scaled by the factor (Np/Ns)^2.

This equivalent is not so complex that it cannot be solved with a pencil. With a circuit simulator it is a snap.

No fiction to it at all. It is an equivalent circuit with its advantages and limitations. We must be aware of the limitations when we use it.
 
Re: Apparent Power, KVA?

No fiction to it at all.
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Huh?

Plug a real world source onto the two left hand terminals and what does the source "see"?

Plug a real world source onto the input terminals (two terminals - apples to apples ;) ) of a real transformer and what does the source "see"?
 
Re: Apparent Power, KVA?

Al,

Just what are you trying to say? You have been sniping at this circuit since I first mentioned it, but so far you have not come up with anything other than field measurements of reflected load impedances which are certainly useful but are no substitute for this equivalent.

I think we all understand the limitations of models. If we don't, then we should get a job at Burger King.

Now I think the typical facilities engineer or PE would not use this equivalent even if he knows about it. And, I doubt that the manufacturers even publish a set of parameters for the model.

Instead, this model would likely be used by transformer designers, and they probably have something more sophisticated today.

For the rest of us, it helps us understand the workings of a transformer and provides some lab experiments for students.

Come on Al, I challenge you to come up with something better. You have until next year to do it!
 
Re: Apparent Power, KVA?

Al
You say ' we should be aware of the limitations when we use it', this is true.
When these models are made they are as accurate as the data inputted into the models, GIGO still applies even to transformer/motor models.
As rattus pointed out these models can be done on a small HP calculator or even with a slide rule without using trig/log tables, although a knowledge of Eulers formulas is helpful. The difference I suppose is the finess method and the brute force method.
But there is no fiction about the equivalent circuit this is why it is called an equivalent circuit, the quantitities represented in the diagram represent actual physical properties of the transformer. The accuracy is as good as the inputted data, taken from a transformer Open circuit test and the short-circuit test.

Gary
 
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